How to solve this puzzle by using Axiom of Choice? [duplicate]
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This question already has an answer here:
Why does the infinite prisoners and hats puzzle require the axiom of choice?
3 answers
In this article, at the end of page 6, it is given the following puzzle,
An evil wizard has threatened a village where an infinite number of
gnomes reside. The wizard will cast a spell that will cause a hat to
appear on the head of every gnome. Each hat will either be red or
blue, but each gnome will be unable to see that hat on his or her
head. The wizard will leave the gnomes alone only if only a finite
number of gnomes guess the color of the hat on their heads
incorrectly. The gnomes can strategize before the wizard puts the hats
on their heads, but they cannot talk or communicate with each other
once the hats are on their heads. The gnomes have very good eyesight
and can see the hat of every other gnome. The wizard can listen to the
gnomes strategize and choose the most evil possible placement of hats.
What should the gnomes do?
[solution, if you can find it, requires the Axiom of Choice.]
I have been thinking this puzzle since yesterday, but I couldn't come up with any solution, so what is best strategy(ies) that can gnomes choose so that the wizard will leave them alone ?
Edit:
Note that there are infinitely many gnomes, not necessarily countably infinite.Morever, the puzzle does not say that a gnome that guesses his/her hat wrong will be kill right away, so there is no way s/he can learn whether his/her choice was true or not before everyone else made their guess.
soft-question recreational-mathematics puzzle axiom-of-choice
asked Jan 8 at 18:42
onurcanbektasonurcanbektas
3,47511037
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Jan 8 at 18:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why does the infinite prisoners and hats puzzle require the axiom of choice?
3 answers
In this article, at the end of page 6, it is given the following puzzle,
An evil wizard has threatened a village where an infinite number of
gnomes reside. The wizard will cast a spell that will cause a hat to
appear on the head of every gnome. Each hat will either be red or
blue, but each gnome will be unable to see that hat on his or her
head. The wizard will leave the gnomes alone only if only a finite
number of gnomes guess the color of the hat on their heads
incorrectly. The gnomes can strategize before the wizard puts the hats
on their heads, but they cannot talk or communicate with each other
once the hats are on their heads. The gnomes have very good eyesight
and can see the hat of every other gnome. The wizard can listen to the
gnomes strategize and choose the most evil possible placement of hats.
What should the gnomes do?
[solution, if you can find it, requires the Axiom of Choice.]
I have been thinking this puzzle since yesterday, but I couldn't come up with any solution, so what is best strategy(ies) that can gnomes choose so that the wizard will leave them alone ?
Edit:
Note that there are infinitely many gnomes, not necessarily countably infinite.Morever, the puzzle does not say that a gnome that guesses his/her hat wrong will be kill right away, so there is no way s/he can learn whether his/her choice was true or not before everyone else made their guess.
soft-question recreational-mathematics puzzle axiom-of-choice
asked Jan 8 at 18:42
onurcanbektasonurcanbektas
3,47511037
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marked as duplicate by José Carlos Santos, Asaf Karagila♦ axiom-of-choice
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Jan 8 at 18:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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And the article José's link references is this: en.wikipedia.org/wiki/…
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– T. Fo
Jan 8 at 18:51
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@T.Ford but the puzzle does not say that there are countable infinitely many gnomes.
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– onurcanbektas
Jan 8 at 18:55
1
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@onurcanbektas The countability requirement therein comes from the prisoners only being able to see in one direction, but each of your gnomes sees every other gnome, so they can be uncountable without harming the theorem.
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– J.G.
Jan 8 at 19:12
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@J.G. which theorem are you referring to ?
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– onurcanbektas
Jan 8 at 19:13
1
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@onurcanbektas That the gnomes win.
$endgroup$
– J.G.
Jan 8 at 19:23
add a comment |
$begingroup$
This question already has an answer here:
Why does the infinite prisoners and hats puzzle require the axiom of choice?
3 answers
In this article, at the end of page 6, it is given the following puzzle,
An evil wizard has threatened a village where an infinite number of
gnomes reside. The wizard will cast a spell that will cause a hat to
appear on the head of every gnome. Each hat will either be red or
blue, but each gnome will be unable to see that hat on his or her
head. The wizard will leave the gnomes alone only if only a finite
number of gnomes guess the color of the hat on their heads
incorrectly. The gnomes can strategize before the wizard puts the hats
on their heads, but they cannot talk or communicate with each other
once the hats are on their heads. The gnomes have very good eyesight
and can see the hat of every other gnome. The wizard can listen to the
gnomes strategize and choose the most evil possible placement of hats.
What should the gnomes do?
[solution, if you can find it, requires the Axiom of Choice.]
I have been thinking this puzzle since yesterday, but I couldn't come up with any solution, so what is best strategy(ies) that can gnomes choose so that the wizard will leave them alone ?
Edit:
Note that there are infinitely many gnomes, not necessarily countably infinite.Morever, the puzzle does not say that a gnome that guesses his/her hat wrong will be kill right away, so there is no way s/he can learn whether his/her choice was true or not before everyone else made their guess.
soft-question recreational-mathematics puzzle axiom-of-choice
asked Jan 8 at 18:42
onurcanbektasonurcanbektas
3,47511037
$endgroup$
This question already has an answer here:
Why does the infinite prisoners and hats puzzle require the axiom of choice?
3 answers
In this article, at the end of page 6, it is given the following puzzle,
An evil wizard has threatened a village where an infinite number of
gnomes reside. The wizard will cast a spell that will cause a hat to
appear on the head of every gnome. Each hat will either be red or
blue, but each gnome will be unable to see that hat on his or her
head. The wizard will leave the gnomes alone only if only a finite
number of gnomes guess the color of the hat on their heads
incorrectly. The gnomes can strategize before the wizard puts the hats
on their heads, but they cannot talk or communicate with each other
once the hats are on their heads. The gnomes have very good eyesight
and can see the hat of every other gnome. The wizard can listen to the
gnomes strategize and choose the most evil possible placement of hats.
What should the gnomes do?
[solution, if you can find it, requires the Axiom of Choice.]
I have been thinking this puzzle since yesterday, but I couldn't come up with any solution, so what is best strategy(ies) that can gnomes choose so that the wizard will leave them alone ?
Edit:
Note that there are infinitely many gnomes, not necessarily countably infinite.Morever, the puzzle does not say that a gnome that guesses his/her hat wrong will be kill right away, so there is no way s/he can learn whether his/her choice was true or not before everyone else made their guess.
This question already has an answer here:
Why does the infinite prisoners and hats puzzle require the axiom of choice?
3 answers
soft-question recreational-mathematics puzzle axiom-of-choice
soft-question recreational-mathematics puzzle axiom-of-choice
asked Jan 8 at 18:42
onurcanbektasonurcanbektas
3,47511037
asked Jan 8 at 18:42
onurcanbektasonurcanbektas
3,47511037
edited Jan 8 at 18:54
onurcanbektas
asked Jan 8 at 18:42
onurcanbektasonurcanbektas
3,47511037
asked Jan 8 at 18:42
onurcanbektasonurcanbektas
3,47511037
asked Jan 8 at 18:42
onurcanbektasonurcanbektas
3,47511037
3,47511037
marked as duplicate by José Carlos Santos, Asaf Karagila♦ axiom-of-choice
Users with the axiom-of-choice badge can single-handedly close axiom-of-choice questions as duplicates and reopen them as needed.
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Jan 8 at 18:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, Asaf Karagila♦ axiom-of-choice
Users with the axiom-of-choice badge can single-handedly close axiom-of-choice questions as duplicates and reopen them as needed.
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Jan 8 at 18:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
And the article José's link references is this: en.wikipedia.org/wiki/…
$endgroup$
– T. Fo
Jan 8 at 18:51
$begingroup$
@T.Ford but the puzzle does not say that there are countable infinitely many gnomes.
$endgroup$
– onurcanbektas
Jan 8 at 18:55
1
$begingroup$
@onurcanbektas The countability requirement therein comes from the prisoners only being able to see in one direction, but each of your gnomes sees every other gnome, so they can be uncountable without harming the theorem.
$endgroup$
– J.G.
Jan 8 at 19:12
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@J.G. which theorem are you referring to ?
$endgroup$
– onurcanbektas
Jan 8 at 19:13
1
$begingroup$
@onurcanbektas That the gnomes win.
$endgroup$
– J.G.
Jan 8 at 19:23
add a comment |
$begingroup$
And the article José's link references is this: en.wikipedia.org/wiki/…
$endgroup$
– T. Fo
Jan 8 at 18:51
$begingroup$
@T.Ford but the puzzle does not say that there are countable infinitely many gnomes.
$endgroup$
– onurcanbektas
Jan 8 at 18:55
1
$begingroup$
@onurcanbektas The countability requirement therein comes from the prisoners only being able to see in one direction, but each of your gnomes sees every other gnome, so they can be uncountable without harming the theorem.
$endgroup$
– J.G.
Jan 8 at 19:12
$begingroup$
@J.G. which theorem are you referring to ?
$endgroup$
– onurcanbektas
Jan 8 at 19:13
1
$begingroup$
@onurcanbektas That the gnomes win.
$endgroup$
– J.G.
Jan 8 at 19:23
$begingroup$
And the article José's link references is this: en.wikipedia.org/wiki/…
$endgroup$
– T. Fo
Jan 8 at 18:51
$begingroup$
And the article José's link references is this: en.wikipedia.org/wiki/…
$endgroup$
– T. Fo
Jan 8 at 18:51
$begingroup$
@T.Ford but the puzzle does not say that there are countable infinitely many gnomes.
$endgroup$
– onurcanbektas
Jan 8 at 18:55
$begingroup$
@T.Ford but the puzzle does not say that there are countable infinitely many gnomes.
$endgroup$
– onurcanbektas
Jan 8 at 18:55
1
1
$begingroup$
@onurcanbektas The countability requirement therein comes from the prisoners only being able to see in one direction, but each of your gnomes sees every other gnome, so they can be uncountable without harming the theorem.
$endgroup$
– J.G.
Jan 8 at 19:12
$begingroup$
@onurcanbektas The countability requirement therein comes from the prisoners only being able to see in one direction, but each of your gnomes sees every other gnome, so they can be uncountable without harming the theorem.
$endgroup$
– J.G.
Jan 8 at 19:12
$begingroup$
@J.G. which theorem are you referring to ?
$endgroup$
– onurcanbektas
Jan 8 at 19:13
$begingroup$
@J.G. which theorem are you referring to ?
$endgroup$
– onurcanbektas
Jan 8 at 19:13
1
1
$begingroup$
@onurcanbektas That the gnomes win.
$endgroup$
– J.G.
Jan 8 at 19:23
$begingroup$
@onurcanbektas That the gnomes win.
$endgroup$
– J.G.
Jan 8 at 19:23
add a comment |
1 Answer
1
active
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Call two ways to colour the hats equivalent if they differ in only finitely many places. Next, form a choice function on the set of such equivalence classes. If each gnome assumes the colouring chosen from their equivalence class is correct, this guess will differ from the true one in at most one place, so is equivalent to it. Since they all work from the same hypothesis, only finitely many are wrong.
answered Jan 8 at 18:53
J.G.J.G.
33.5k23252
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Call two ways to colour the hats equivalent if they differ in only finitely many places. Next, form a choice function on the set of such equivalence classes. If each gnome assumes the colouring chosen from their equivalence class is correct, this guess will differ from the true one in at most one place, so is equivalent to it. Since they all work from the same hypothesis, only finitely many are wrong.
answered Jan 8 at 18:53
J.G.J.G.
33.5k23252
$endgroup$
add a comment |
$begingroup$
Call two ways to colour the hats equivalent if they differ in only finitely many places. Next, form a choice function on the set of such equivalence classes. If each gnome assumes the colouring chosen from their equivalence class is correct, this guess will differ from the true one in at most one place, so is equivalent to it. Since they all work from the same hypothesis, only finitely many are wrong.
answered Jan 8 at 18:53
J.G.J.G.
33.5k23252
$endgroup$
add a comment |
$begingroup$
Call two ways to colour the hats equivalent if they differ in only finitely many places. Next, form a choice function on the set of such equivalence classes. If each gnome assumes the colouring chosen from their equivalence class is correct, this guess will differ from the true one in at most one place, so is equivalent to it. Since they all work from the same hypothesis, only finitely many are wrong.
answered Jan 8 at 18:53
J.G.J.G.
33.5k23252
$endgroup$
Call two ways to colour the hats equivalent if they differ in only finitely many places. Next, form a choice function on the set of such equivalence classes. If each gnome assumes the colouring chosen from their equivalence class is correct, this guess will differ from the true one in at most one place, so is equivalent to it. Since they all work from the same hypothesis, only finitely many are wrong.
answered Jan 8 at 18:53
J.G.J.G.
33.5k23252
answered Jan 8 at 18:53
J.G.J.G.
33.5k23252
answered Jan 8 at 18:53
J.G.J.G.
33.5k23252
answered Jan 8 at 18:53
J.G.J.G.
33.5k23252
33.5k23252
add a comment |
add a comment |
$begingroup$
And the article José's link references is this: en.wikipedia.org/wiki/…
$endgroup$
– T. Fo
Jan 8 at 18:51
$begingroup$
@T.Ford but the puzzle does not say that there are countable infinitely many gnomes.
$endgroup$
– onurcanbektas
Jan 8 at 18:55
1
$begingroup$
@onurcanbektas The countability requirement therein comes from the prisoners only being able to see in one direction, but each of your gnomes sees every other gnome, so they can be uncountable without harming the theorem.
$endgroup$
– J.G.
Jan 8 at 19:12
$begingroup$
@J.G. which theorem are you referring to ?
$endgroup$
– onurcanbektas
Jan 8 at 19:13
1
$begingroup$
@onurcanbektas That the gnomes win.
$endgroup$
– J.G.
Jan 8 at 19:23